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Question
Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant
The other factor in the value of the determinant is
Options
4
2
a + b + c
none of these
Solution
\[\Delta = \begin{vmatrix} - 2a & a + b & a + c\\b + a & - 2b & b + c\\c + a & c + b & - 2c \end{vmatrix}
\text{ Let } a + b = 2C, b + c = 2A\text{ and }c + a = 2B . \]
\[ \Rightarrow a + b + b + c + c + a = 2A + 2B + 2C\]
\[ \Rightarrow 2\left( a + b + c \right) = 2\left( A + B + C \right)\]
\[ \Rightarrow a + b + c = A + B + C\]
Also,
\[a = \left( a + b + c \right) - \left( b + c \right) = \left( A + B + C \right) - 2A = B + C - A\]
Similarly,
\[b = C + A - B, c = A + B - C\]
\[\Delta = \begin{vmatrix} 2A - 2B - 2C & 2C & 2B\\ 2C & 2B - 2C - 2A & 2A\\ 2B & 2A & 2C - 2A - 2B \end{vmatrix} = 8 \times \begin{vmatrix} A - B - C & C & B\\ C & B - C - A & A\\ B & A & C - A - B \end{vmatrix} \left[\text{ taking out 2 common from }R_1 R_2 R_3 \right]\]
\[ = 8 \times \begin{vmatrix} A - B & C + B & B\\ B - A & B - C & A\\ B + A & C - B & C - A - B \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 + C_2 , C_2 \to C_2 + C_3 \right]\]
\[ = 8 \times \begin{vmatrix} \left( A - B \right) & C + B & B\\ 0 & 2B & A + B\\ 2B & 0 & C - B \end{vmatrix} \left[\text{ Applying }R_2 \to R_1 + R_2 , R_3 \to R_2 + R_3 \right]\]
\[ = 8 \times \left\{ \left( A - B \right)\begin{vmatrix} 2B & A + B\\ 0 & C - B \end{vmatrix} + \left( 2B \right) \times \begin{vmatrix} C + B & B\\ 2B & A + B \end{vmatrix} \right\} \left[\text{ Expanding along }C_1 \right]\]
\[ = 16 B\left\{ \left( A - B \right)\left( C - B \right) + \left( C + B \right)\left( A + B \right) - 2 B^2 \right\}\]
\[ = 32 ABC\]
\[ = 32\left( \frac{b + c}{2} \right)\left( \frac{c + a}{2} \right)\left( \frac{a + b}{2} \right)\]
\[ = 4\left( a + b \right)\left( b + c \right)\left( c + a \right)\]
Hence, 4 is the other factor of the determinant .
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